2003 AMC 12B Problems/Problem 17: Difference between revisions

Revision as of 18:54, 29 September 2014

Problem

If $\log (xy^3) = 1$ and $\log (x^2y) = 1$ , what is $\log (xy)$ ?

$\mathrm{(A)}\ -\frac 12 \qquad\mathrm{(B)}\ 0 \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 35 \qquad\mathrm{(E)}\ 1$

Solution

Since $\begin{align*} &\log(xy) +2\log y = 1 \\ \log(xy) + \log x = 1 \quad \Longrightarrow \quad &2\log(xy) + 2\log x = 2 \end{align*}$ Summing gives $3\log(xy) + 2\log y + 2\log x = 3 \Longrightarrow 5\log(xy) = 3$

Hence $\log (xy) = \frac 35 \Rightarrow \mathrm{(D)}$ .

It is not difficult to find $x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}$ .

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

@@ Line 18: / Line 18: @@
 Hence <math>\log (xy) = \frac 35 \Rightarrow \mathrm{(D)}</math>.
-It is not difficult to find <math>x = 10^{2/5}, y = 10^{1/5}</math>.
+It is not difficult to find <math>x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}</math>.
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2003 AMC 12B Problems/Problem 17: Difference between revisions

Revision as of 18:54, 29 September 2014

Problem

Solution

See also